如何在Python中反向迭代列表?
请参见:如何获得列表的反向副本(避免在.reverse后链接方法时使用单独的语句)?
如何在Python中反向迭代列表?
请参见:如何获得列表的反向副本(避免在.reverse后链接方法时使用单独的语句)?
当前回答
The cute Solution
class List_reverse():
def reverse_list(self):
list = [1, 2, 3, 4, 5, 6]
list1 = []
for i in range(1, len(list)+1):
list1.append(list[-i])
print(list1)
if __name__ == "__main__":
obj = List_reverse()
obj.reverse_list()
enter code here
其他回答
>>> L = [1, 2, 3, 4]
>>> L = [L[-i] for i in range(1, len(L) + 1)]
>>> L
[4, 3, 2, 1]
另一种方法使用列表理解和abs
array = [0, 10, 20, 40]
reversed_array = [array[abs(indx)] for indx in range(abs(len(array)-1),1)]
reversed_array
[40, 20, 10, 0]
带颠倒和列表的:
>>> list1 = [1,2,3]
>>> reversed_list = list(reversed(list1))
>>> reversed_list
>>> [3, 2, 1]
这个类使用Python魔法方法和迭代器进行反转,并反转列表:
class Reverse(object):
""" Builds a reverse method using magic methods """
def __init__(self, data):
self.data = data
self.index = len(data)
def __iter__(self):
return self
def __next__(self):
if self.index == 0:
raise StopIteration
self.index = self.index - 1
return self.data[self.index]
REV_INSTANCE = Reverse([0, 10, 20, 40])
iter(REV_INSTANCE)
rev_list = []
for i in REV_INSTANCE:
rev_list.append(i)
print(rev_list)
输出
[40, 20, 10, 0]
使用一些逻辑
用一些老派的逻辑来练习面试。
从前到后交换数字。使用两个指针索引[0]和索引[last]
def reverse(array):
n = array
first = 0
last = len(array) - 1
while first < last:
holder = n[first]
n[first] = n[last]
n[last] = holder
first += 1
last -= 1
return n
input -> [-1 ,1, 2, 3, 4, 5, 6]
output -> [6, 5, 4, 3, 2, 1, -1]